Δd ball / Force Applied Ball - Completely stumped.

  1. Monocerotis

    Monocerotis 50
    Gold Member

    1. The problem statement, all variables and given/known data

    There are two short questions I'm absolutely stumped by. And I can't even find a way to work backwards from the answer to understand them.

    Please help if you can, these two questions are driving me nuts ;__;

    Question 1)
    by applying a force you accelerate a ball upwards at 16m/s2 for 0.5 seconds. How high does the ball go after you let go (from it's starting position)?
    --> Answer: 5.26m

    Question 2)
    How much applied force is required to accelerate a 2.0kg ball upwards at 3.0 m/s2
    --> Answer: 27N [Up]

    2. Relevant equations
    Ref. 3.
    3. The attempt at a solution

    Question 1 Attempt
    V2= V1(Δt)+ 1/2 Aav(Δt)2
    V2= 0(0.5)+ 16m/s2(1/4)
    V2 = 4m/s

    Δd = V2(Δt) - 1/2 Aav(Δt)2
    Δd = 4m/s(0.5s) - 1/2 16m/s(0.5s)2
    Δd = 2 - 2
    Δd = 0 ?????????

    Question 2 Attempt
    Fnet1=(m)(a)
    Fnet1 = 2.0kg(3.0m/s2)
    Fnet1 = 6N

    Fnet2 = (m)(a = g)
    Fnet2 = (2.0kg)(9.8m/s2)
    Fnet2 = 19.6N

    Fnet total = 19.6 + 6
    Fnet total = 25.6N
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Chegg
    EDIT: I had to edit this, the output came out completely wrong...
    EDIT 2: edited again, sorry if this doesn't show up right... physicsforums is strange to use with tex code
    EDIT 3: found the problem, i wish physics forums would explain the $ sign is not necessary to write for tex code math here

    Question 1:

    Okay, one thing [itex]\Delta d[/itex] is wrong because you don't get [itex]\Delta t ^2[/itex], but [itex] t_f ^2 - t_i^2[/itex] which is not the same (this is important). Your reasoning is right other than that, and I would try to just find the time where the velocity is zero and plug that in the distance equation so (as usual):

    [itex]v ( \tau ) = v2 - gt = 0[/itex]

    so:

    [itex] \tau = \frac{v2}{g}[/itex]

    then plug that in you kinematic equation for distance:

    [itex]x(\tau) = x_0 + v_i \tau - \frac{1}{2} g \tau^2[/itex]

    And find [itex]x(\tau)[/itex] with the values you have.

    I'm sure you've already figured this part out before I wrote it ;-)

    Question 2.

    Your answer is right, I don't know why they put 27N.

    But how you did it I'm not quite sure so I'll write an explanation anyway to convince you more your answer is right.

    You have gravity in this situation, so draw your force body diagram.
    So you should draw it and get:

    [itex]F_{net} = \vec{F}_{applied} + \vec{F}_{gravity}[/itex]

    looking at the directions (gravity is down, net accel is up, and your force obviously has to be up), you get:

    [itex]m a_{net} = F_{applied} - mg[/itex]

    so:

    [itex]F_{applied} = m (a_{net} + g)[/itex]

    When you look at forces, what you have to do is look at everything that is acting on the body and add them up (thanks to superposition, this is a very strong and very important idea in physics, it would be good to spend time to think about how powerful it is :-))

    What this problem was asking is what force can you add to make the outcome of the acceleration [itex]3.0 \frac{m}{s^2}[/itex], and NOT, what is the outcome of accelerating a mass at [itex]3.0 \frac{m}{s^2}[/itex] and then adding gravity's acceleration to the situation (so the accelerations subtract).


    If anything is confusing or weird please let me know I hope this helps :-)
     
    Last edited: Oct 18, 2009
  4. Monocerotis

    Monocerotis 50
    Gold Member

    Thanks for the help ordirules, but could you define yours symbols ?

    I'm not sure what [itex]
    v ( \tau ) = v2 - gt = 0
    [/itex]

    [itex]
    x(\tau) = x_0 + v_i \tau - \frac{1}{2} g \tau^2
    [/itex]

    [itex]
    \tau = \frac{v2}{g}
    [/itex]

    are supposed to mean.

    :S lol
     
  5. [itex]

    v ( \tau ) = v2 - gt = 0

    [/itex]
    This means the speed v is equal to the speed of the particle v2 once it has stopped accelerating minus the accel of gravity times time. This is a kinematic equation which I'm sure you have in your text.
    If you know calculus, these are super easy to derive too but I won't get into that :-).

    [itex]

    x(\tau) = x_0 + v_i \tau - \frac{1}{2} g \tau^2

    [/itex]
    This equation is a kinematic equation that I'm sure your textbook has. What you need to do with this equation is plug in the time that you get by solving the other equation higher above for when the velocity is equal to zero.

    [tex]\tau[/tex] is a greek symbol commonly used instead of the letter "t". I used it just to try to emphasize that the letter is representing a numerical value because you solved for it. In that last equation, you just solved the equation at the top of this reply for v = 0.

    Sorry for the confusion, I hope this helps, good luck!
     
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